Class 9 Maths Chapter 10 How Quantities Combine: Understanding Data notes, NCERT solutions, easy concepts, solved questions and practice worksheet based on the new Ganita Manjari textbook 2026–27.
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Chapter 10 – How Quantities Combine: Understanding Data
10.1 Combining Things
10.1.1 Average of Averages
Easy Concept:
When two or more groups are combined, we cannot always simply take the average of their averages. The size of each group must also be considered.
Combined Average
Combined average =
(Sum of all values) ÷ (Total number of values)
Combined average =
(Sum of all values) ÷ (Total number of values)
If Collection 1 has n values and average a, its total is an.
If Collection 2 has m values and average b, its total is bm.
Combined average =
(an + bm) ÷ (n + m)
Remember:
The average of averages can be found by simply adding the averages and dividing by the number of groups only when all groups have the same size.
Example 1
In the badminton academy example, there are 8 seniors with average height 165.5 cm and 3 juniors with average height 149.33 cm.
Total height of seniors:
165.5 × 8 = 1324 cm
Total height of juniors:
149.33 × 3 ≈ 448 cm
Therefore,
Average height =
(1324 + 448) ÷ 11
≈ 161.09 cm
≈ 161.09 cm
So the average height of the whole group is approximately 161.09 cm.
Example 2 – Cycling
Jaspreet cycled for 5 weekdays in each of 3 weeks. The weekly averages were 12.8 minutes, 15.8 minutes and 18 minutes.
Since every week has the same number of cycling days, each weekly average has equal weight.
Combined average =
(12.8 + 15.8 + 18) ÷ 3
= 46.6 ÷ 3
≈ 15.53 minutes
= 46.6 ÷ 3
≈ 15.53 minutes
Therefore, the average time spent cycling per weekday is 15.53 minutes.
Important Rule:
If all collections have equal sizes, the combined average is simply the average of the individual averages. If their sizes are different, use a weighted average.
10.1.2 Mixtures
Easy Concept:
When different mixtures are combined, the final concentration depends on both:
- the concentration of each mixture, and
- the quantity of each mixture.
Concentration of mixture =
Total quantity of substance ÷ Total quantity of mixture
Total quantity of substance ÷ Total quantity of mixture
Example 3
Two equal quantities of lemonade contain 10% and 20% jaggery.
Because the quantities are equal, the final concentration is halfway between 10% and 20%.
(10 + 20) ÷ 2 = 15%
Answer: 15% jaggery.
Example 4
A bowl contains 500 mL lemonade with 10% jaggery. A glass contains 200 mL lemonade with 20% jaggery.
Jaggery in bowl:
500 × 0.10 = 50 mL
Jaggery in glass:
200 × 0.20 = 40 mL
Total jaggery = 90 mL
Total lemonade = 700 mL
Concentration =
90 ÷ 700 × 100
≈ 12.86%
≈ 12.86%
Answer: Approximately 12.86%.
Example 5 – Brass Alloy
Batch A = 200 kg, 70% copper.
Batch B = 120 kg, 50% copper.
Batch C = y kg, 45% copper.
Final mixture = 55% copper.
Copper in A: 200 × 0.70 = 140 kg
Copper in B: 120 × 0.50 = 60 kg
Using the concentration equation:
0.55(320 + y) = 140 + 60 + 0.45y
320 × 0.55 + 0.55y = 200 + 0.45y
176 + 0.55y = 200 + 0.45y
0.10y = 24
y = 240 kg
Therefore, Batch C weighs 240 kg.
Weighted Mean
Weighted mean =
(w₁x₁ + w₂x₂ + w₃x₃ + … + wₙxₙ) ÷ (w₁ + w₂ + w₃ + … + wₙ)
(w₁x₁ + w₂x₂ + w₃x₃ + … + wₙxₙ) ÷ (w₁ + w₂ + w₃ + … + wₙ)
In simple words:
A weighted average gives more importance to values that represent larger quantities or have greater importance.
Historical Note
The weighted mean was described by Indian mathematician Brahmagupta in Brahmasphutasiddhanta around 628 CE. The idea was also used by Indian mathematicians such as Sridharacharya for problems involving purity of gold alloys.
Book Questions – Section 10.1
1. The average score of students on a test in Section A is 72 and that of students in Section B is 76. What is the combined average of both the sections given that Section A has: 30 students; and Section B has 25 students?
Total score of Section A:
72 × 30 = 2160
Total score of Section B:
76 × 25 = 1900
Total score = 2160 + 1900 = 4060
Total students = 30 + 25 = 55
Combined average = 4060 ÷ 55
≈ 73.82
Answer: Approximately 73.82.
72 × 30 = 2160
Total score of Section B:
76 × 25 = 1900
Total score = 2160 + 1900 = 4060
Total students = 30 + 25 = 55
Combined average = 4060 ÷ 55
≈ 73.82
Answer: Approximately 73.82.
2. A farmer mixes three equal quantities of fertilisers. The first one contains 1/10 nitrogen, the second contains 9/50 nitrogen, and the third contains 3/60 nitrogen. What is the fraction of nitrogen in the mixture?
three equal quantities of fertilisers are mixed.
the fractions of nitrogen are
1/10, 9/50, 3/60
the fraction of nitrogen in the mixture in their average
= (1/10 + 9/50 + 3/60) ÷ 3
= (30/300 + 54/300 + 15 / 300) ÷ 3
= 99/300 ÷ 3
= 99/900
= 11/100
Answer: the fraction of nitrogen in the mixture is 11/100 .
the fractions of nitrogen are
1/10, 9/50, 3/60
the fraction of nitrogen in the mixture in their average
= (1/10 + 9/50 + 3/60) ÷ 3
= (30/300 + 54/300 + 15 / 300) ÷ 3
= 99/300 ÷ 3
= 99/900
= 11/100
Answer: the fraction of nitrogen in the mixture is 11/100 .
3. (Sridharacharya, Patiganita, c.750 CE) In ancient India, Varna was the measure of gold purity. A purity of 16 Varna meant pure gold; in general, a purity of k varna meant that the gold alloy was k/16 gold and the rest impurities. (Now the form used is karat; 16 Varna = 24 karat.) Suppose a goldsmith melts together three pieces of gold: 9 units at 12 varna, 5 units at 10 varna, and 17 units at 11 varna. Find the purity in varna of the combined gold.
Use the weighted average:
Purity = (9 × 12 + 5 × 10 + 17 × 11) ÷ (9 + 5 + 17)
= (1088 + 50 + 187) ÷ 31
= 345 ÷ 31
= 11.129 varna
Answer: 11.13 varna.
Purity = (9 × 12 + 5 × 10 + 17 × 11) ÷ (9 + 5 + 17)
= (1088 + 50 + 187) ÷ 31
= 345 ÷ 31
= 11.129 varna
Answer: 11.13 varna.
4. The average rainfall per day in the months of May, June, and July in a certain location are 3.5 mm, 10 mm and 8.7 mm respectively. Write an expression that gives their combined average.
There are three months and each average represents one month.
Therefore the required expression is:
(3.5 + 10 + 8.7) ÷ 3
= 22.2 ÷ 3
= 7.4 mm
Answer: 7.4 mm.
(3.5 + 10 + 8.7) ÷ 3
= 22.2 ÷ 3
= 7.4 mm
Answer: 7.4 mm.
5. Calculate the concentration of spice mix in these two scenarios. (i) A 100 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 300 mL one with 15% spice mix are combined. (ii) A 300 mL kashayam/kadha with 5% spice mix, a 200 mL one with 10% spice mix, and a 100 mL one with 15% spice mix are mixed.
(i)
Spice = 100 × 0.05 + 200 × 0.10 + 300 × 0.15
= 5 + 20 + 45 = 70 mL
Total mixture = 600 mL
Concentration = 70/600 × 100
= 11.67%
(ii)
Spice = 300 × 0.05 + 200 × 0.10 + 100 × 0.15
= 15 + 20 + 15 = 50 mL
Total mixture = 600 mL
Concentration = 50/600 × 100
= 8.33%
Answers: (i) 11.67% approximately; (ii) 8.33% approximately.
Spice = 100 × 0.05 + 200 × 0.10 + 300 × 0.15
= 5 + 20 + 45 = 70 mL
Total mixture = 600 mL
Concentration = 70/600 × 100
= 11.67%
(ii)
Spice = 300 × 0.05 + 200 × 0.10 + 100 × 0.15
= 15 + 20 + 15 = 50 mL
Total mixture = 600 mL
Concentration = 50/600 × 100
= 8.33%
Answers: (i) 11.67% approximately; (ii) 8.33% approximately.
10.1.3 Custom Weights
Easy Concept:
Sometimes different parts of data have different importance. We then assign weights to them.
Example 6
Rehmat scored:
- Internal tests = 60%
- Project = 64%
- Final exam = 73%
The weights are 3 : 2 : 5.
Annual score =
(60 × 3 + 64 × 2 + 73 × 5) ÷ (3 + 2 + 5)
= (180 + 128 + 365) ÷ 10
= 673 ÷ 10
= 67.3%
Book Questions – Section 10.2
1. Savitri’s marks in Kashmiri are as follows: 35 out of 50 in internal tests, 44 out of 60 in the project, and 80 out of 100 in the final exam. The annual percentage score is calculated by combining the internals, project, and final exam in the ratio 3:4:5. Which of the following expression(s) gives her annual score (as a percentage) in Kashmiri?
Convert each mark into a percentage first:
Internal tests: 35/50 × 100 = 70%
Project: 44/60 × 100 = 73.33%
Final exam: 80/100 × 100 = 80%
The weighted average is:
(70 × 3 + 73.33 × 4 + 80 × 5) ÷ (3 + 4 + 5)
So the correct expression is the one that first converts each score into a percentage and then applies weights 3, 4 and 5.
Answer: (iii) and (iv) are correct.
Internal tests: 35/50 × 100 = 70%
Project: 44/60 × 100 = 73.33%
Final exam: 80/100 × 100 = 80%
The weighted average is:
(70 × 3 + 73.33 × 4 + 80 × 5) ÷ (3 + 4 + 5)
So the correct expression is the one that first converts each score into a percentage and then applies weights 3, 4 and 5.
Answer: (iii) and (iv) are correct.
Where weighted averages are used:
- Restaurant ratings
- Product ratings
- Examination marks
- Mixtures
- Gold purity
- Average prices
- Scientific data
Book Questions – Section 10.3
1. A stationery shop owner made ₹8000 selling books, of which 30% is the profit amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the percentage of profit on the total sales?
Profit from books:
8000 × 30/100 = ₹2400
Profit from covers:
1000 × 50/100 = ₹500
Total profit = ₹2900
Total sales = ₹9000
Profit percentage:
2900/9000 × 100
≈ 32.22%
Answer: Approximately 32.22%.
8000 × 30/100 = ₹2400
Profit from covers:
1000 × 50/100 = ₹500
Total profit = ₹2900
Total sales = ₹9000
Profit percentage:
2900/9000 × 100
≈ 32.22%
Answer: Approximately 32.22%.
2. A white stork’s migration is tracked. The average daily distance travelled, calculated over 20 days, is 44.5 km. On the 21st day, it flew 55 km. What is the average daily distance travelled over these 21 days? Make a guess before you calculate.
Distance in first 20 days:
20 × 44.5 = 890 km
Distance on 21st day = 55 km
Total = 890 + 55 = 945 km
Average = 945 ÷ 21
≈ 45 km
Answer: 45 km per day.
20 × 44.5 = 890 km
Distance on 21st day = 55 km
Total = 890 + 55 = 945 km
Average = 945 ÷ 21
≈ 45 km
Answer: 45 km per day.
3. A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar. What are the concentrations of salt and sugar in the mixture?
Salt:
600 × 5/100 = 30 mL
Total mixture = 900 mL
Salt concentration = 30/900 × 100 = 3.33%
Sugar:
300 × 8/100 = 24 mL
Sugar concentration = 24/900 × 100 = 2.67%
Answer: Salt = 3.33%, Sugar = 2.67%.
Therefore the correct option is (v).
600 × 5/100 = 30 mL
Total mixture = 900 mL
Salt concentration = 30/900 × 100 = 3.33%
Sugar:
300 × 8/100 = 24 mL
Sugar concentration = 24/900 × 100 = 2.67%
Answer: Salt = 3.33%, Sugar = 2.67%.
Therefore the correct option is (v).
4. At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water) was 8%. Many customers complained that it was too spicy. What quantity of regular water should be mixed into the 10 litres of pani so that the spice level is reduced to 3/4 of the original concentration?
Original concentration = 8%.
Required concentration = 3/4 × 8% = 6%.
Spice in 10 litres:
10 × 8/100 = 0.8 litre
Let x litres of water be added.
Total mixture = 10 + x
Required concentration:
0.8/(10+x) = 6/100
0.8 = 0.06(10+x)
0.8 = 0.6 + 0.06x
0.2 = 0.06x
x = 10/3 litres
≈ 3.33 litres
Answer: Approximately 3.33 litres of water.
Required concentration = 3/4 × 8% = 6%.
Spice in 10 litres:
10 × 8/100 = 0.8 litre
Let x litres of water be added.
Total mixture = 10 + x
Required concentration:
0.8/(10+x) = 6/100
0.8 = 0.06(10+x)
0.8 = 0.6 + 0.06x
0.2 = 0.06x
x = 10/3 litres
≈ 3.33 litres
Answer: Approximately 3.33 litres of water.
5. A physical fitness evaluation is being undertaken. The final marks are calculated by combining the marks for strength, flexibility, and agility in the ratio 4:5:6. Rashi has scored 60, 65, and 70. Keerthi has scored 55, 65, and 75 respectively. (i) Find out whose total is more without calculating. (ii) What are their final marks?
(i)
Compare the changes:
Rashi: 60, 65, 70
Keerthi: 55, 65, 75
The weights are 4, 5, 6. Rashi is 5 marks higher in strength, equal in flexibility, and 5 marks lower in agility. Since agility has the largest weight, Keerthi’s advantage in agility balances and exceeds Rashi’s advantage in strength. Thus Keerthi’s weighted total is higher.
(ii) Rashi:
(60×4 + 65×5 + 70×6) ÷ 15
= (240 + 325 + 420) ÷ 15
= 985 ÷ 15
= 65.67 approximately.
Keerthi:
(55×4 + 65×5 + 75×6) ÷ 15
= (220 + 325 + 450) ÷ 15
= 995 ÷ 15
= 66.33 approximately.
Answer: Keerthi has the higher final mark.
Compare the changes:
Rashi: 60, 65, 70
Keerthi: 55, 65, 75
The weights are 4, 5, 6. Rashi is 5 marks higher in strength, equal in flexibility, and 5 marks lower in agility. Since agility has the largest weight, Keerthi’s advantage in agility balances and exceeds Rashi’s advantage in strength. Thus Keerthi’s weighted total is higher.
(ii) Rashi:
(60×4 + 65×5 + 70×6) ÷ 15
= (240 + 325 + 420) ÷ 15
= 985 ÷ 15
= 65.67 approximately.
Keerthi:
(55×4 + 65×5 + 75×6) ÷ 15
= (220 + 325 + 450) ÷ 15
= 995 ÷ 15
= 66.33 approximately.
Answer: Keerthi has the higher final mark.
6. A restaurant collected ratings from 10 costomers on a scale of 1to 5. The resulting data is shown in the table below
what is the average rating if the matrics are combined with the weights food :ambience :service = 6:5:4
(i)
the table gives the number f costomers giving each rating
food; average = {5(5) + 3(4) + 2(3) + 0(2) + 0(1)} /10
= (25+12+6)/10 = 43/10 = 4.3
Ambience; Average = {0(5)+4(4) +5(3) +1(2)+ 0(1)} /10
= (16+15+2)/10 =3.3
Service; Average = {1(5)+2(4) +2(3) +4(2)+ 1(1)} /10
=(5+8+6+8+1)/10 = 2.8
The weights are;
Food:Ambience:service=6:5:4
therefore, combined average= {4.3(6)+3.3(5) +2.8(4)}/(6+5+4)
=(25.8+16.5=11.2)/15 = 53.5/15 = 3.57
Answer: 3.57 stars (approx.)
the table gives the number f costomers giving each rating
food; average = {5(5) + 3(4) + 2(3) + 0(2) + 0(1)} /10
= (25+12+6)/10 = 43/10 = 4.3
Ambience; Average = {0(5)+4(4) +5(3) +1(2)+ 0(1)} /10
= (16+15+2)/10 =3.3
Service; Average = {1(5)+2(4) +2(3) +4(2)+ 1(1)} /10
=(5+8+6+8+1)/10 = 2.8
The weights are;
Food:Ambience:service=6:5:4
therefore, combined average= {4.3(6)+3.3(5) +2.8(4)}/(6+5+4)
=(25.8+16.5=11.2)/15 = 53.5/15 = 3.57
Answer: 3.57 stars (approx.)
7. athe following table shows the weight data of langurs in an animal facility. without doing any computations , can yu tell whether there are more male langurs or more female langurs ? or are they equal in number?
(i) Which of the following expression(s) describes the given scenario?
(iii) A female langur weighing 15.2 kg is admitted to the facility. What is the average weight of the female langurs after this?
(iv) Two male langurs weighing 16.9 kg and 16.1 kg are released from the facility. What is the average weight of the male langurs after this?
(v) Now, suppose one of the male langurs lost 1 kg of weight. What is the average weight of all the male langurs after this?
Given:
Average weight of male langurs = 16.5 kg
Average weight of female langurs = 13.8 kg
Average weight of all langurs = 14.925 kg
Total langurs = 60
Let the number of male langurs be x.
Then the number of female langurs is: 60-x
(i) Which expression describes the scenario?
The combined average is: {16.5x+13.8(60-x)}/60 =14.925
So the correct expression is:(a)
(ii) Find the number of male langurs
{16.5x+13.8(60-x)}/60 =14.925
Multiply by 60:
16.5x+13.8(60-x)=14.925(60)
16.5x+828-13.8x=895.5
2.7x=67.5
x=25
Therefore,25 male langurs Female langurs:60-25=35 (iii) A female langur weighing 15.2 kg is admitted
There were 35 female langurs.
Original total weight: 35(13.8)=438kg
After adding the new langur: 483+15.2=498.2 kg
Number of female langurs: 35+1=36
New average: 498.2/36 = 13.838
Therefore,13.84 (approx.)
(iv) Two male langurs weighing 16.9 kg and 16.1 kg are released
Original number of male langurs = 25.
Original total weight: 25(16.5)=412.5 kg
Weight released: 16.9+16.1=33 kg
Remaining total weight: 412.5-33=379.5 kg
Remaining number: 25-2=23
New average:379.5/23=16.5
Therefore,16.5 kg
(v) One male langur loses 1 kg
After part (iv):
Number of male langurs = 23
Total weight = 379.5 kg
One langur loses 1 kg: 379.5-1=378.5 kg
New average: 378.5/23 = 16.4565
Therefore, 16.46 kg (approx.)
Answer:7(i)(a)
7(ii) 25 male langurs
7(iii) 13.84 kg
7(iv) 16.5 kg
7(v) 16.46 kg
Average weight of female langurs = 13.8 kg
Average weight of all langurs = 14.925 kg
Total langurs = 60
Let the number of male langurs be x.
Then the number of female langurs is: 60-x
(i) Which expression describes the scenario?
The combined average is: {16.5x+13.8(60-x)}/60 =14.925
So the correct expression is:(a)
(ii) Find the number of male langurs
{16.5x+13.8(60-x)}/60 =14.925
Multiply by 60:
16.5x+13.8(60-x)=14.925(60)
16.5x+828-13.8x=895.5
2.7x=67.5
x=25
Therefore,25 male langurs Female langurs:60-25=35 (iii) A female langur weighing 15.2 kg is admitted
There were 35 female langurs.
Original total weight: 35(13.8)=438kg
After adding the new langur: 483+15.2=498.2 kg
Number of female langurs: 35+1=36
New average: 498.2/36 = 13.838
Therefore,13.84 (approx.)
(iv) Two male langurs weighing 16.9 kg and 16.1 kg are released
Original number of male langurs = 25.
Original total weight: 25(16.5)=412.5 kg
Weight released: 16.9+16.1=33 kg
Remaining total weight: 412.5-33=379.5 kg
Remaining number: 25-2=23
New average:379.5/23=16.5
Therefore,16.5 kg
(v) One male langur loses 1 kg
After part (iv):
Number of male langurs = 23
Total weight = 379.5 kg
One langur loses 1 kg: 379.5-1=378.5 kg
New average: 378.5/23 = 16.4565
Therefore, 16.46 kg (approx.)
Answer:7(i)(a)
7(ii) 25 male langurs
7(iii) 13.84 kg
7(iv) 16.5 kg
7(v) 16.46 kg
8. Dorjee has collected 1 litre of water from the Dead Sea! Using the information in the table, answer the following questions. A calculator can be used if necessary
(i) What is the salinity of the mixture if he mixes 1 litre of water from the Dead Sea with 2 litres of purified drinking water?
(ii) Is it possible to mix water from the Dead Sea and purified drinking water to get a mixture with the salinity of groundwater? Why/Why not? What quantity of purified drinking water should Dorjee mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of groundwater?
(iii) Is it possible to mix water from the Dead Sea and groundwater to get a mixture with the salinity of purified drinking water? Why/Why not? What quantity of groundwater should he mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of purified drinking water?
(i)
Dead Sea water:1×34=34
Purified drinking water: 2×0.001=0.002
Total salt percentage contribution: 34+0.002=34.002
Total quantity of water:1+2=3 litres
Therefore,
salinity=34.002/3=11.334%
(ii)
Groundwater salinity is:0.01%
Let x litres of purified drinking water be mixed with 1 litre of Dead Sea water.
Then: (34×1+0.001x)/(1+x)=0.01
34+0.001x=0.01+0.01x
33.99=0.009x
x=33.99/0.009
x=3776.67
Therefore, He should mix: 3776.67 litres of purified drinking water with 1 litre of Dead Sea water.
(iii)
Here: Dead Sea water = 34
Groundwater = 0.01
Required salinity =0.001% < 0.01% < 34%
The lowest salinity available among the two waters being mixed is already 0.01.
Therefore, their mixture cannot have a salinity lower than 0.01.
So it is not possible to obtain 0.001.
Therefore, there is no quantity of groundwater that can be mixed with 1 litre of Dead Sea water to obtain the salinity of purified drinking water.
Dead Sea water:1×34=34
Purified drinking water: 2×0.001=0.002
Total salt percentage contribution: 34+0.002=34.002
Total quantity of water:1+2=3 litres
Therefore,
salinity=34.002/3=11.334%
(ii)
Groundwater salinity is:0.01%
Let x litres of purified drinking water be mixed with 1 litre of Dead Sea water.
Then: (34×1+0.001x)/(1+x)=0.01
34+0.001x=0.01+0.01x
33.99=0.009x
x=33.99/0.009
x=3776.67
Therefore, He should mix: 3776.67 litres of purified drinking water with 1 litre of Dead Sea water.
(iii)
Here: Dead Sea water = 34
Groundwater = 0.01
Required salinity =0.001% < 0.01% < 34%
The lowest salinity available among the two waters being mixed is already 0.01.
Therefore, their mixture cannot have a salinity lower than 0.01.
So it is not possible to obtain 0.001.
Therefore, there is no quantity of groundwater that can be mixed with 1 litre of Dead Sea water to obtain the salinity of purified drinking water.
10.2.1 Stacking Columns
What is a stacked bar/column chart?
A stacked bar chart joins different parts of a total into one bar. It allows us to see both the total and the parts making up the total.
10.2 Visualising and Interpreting Data
Table 10.2 – Average Weekly Expenditure
| Family | Housing | Food | Transportation | Healthcare | Education | Recreation |
|---|---|---|---|---|---|---|
| Family A | 1940 | 1700 | 1280 | 1200 | 1770 | 535 |
| Family B | 1750 | 1546 | 1500 | 1280 | 1210 | 0 |
| Family C | 950 | 1700 | 1540 | 1400 | 1300 | 150 |
1. Which family spent the least on housing?
Compare housing:
Family A = ₹1940
Family B = ₹1750
Family C = ₹950
Answer: Family C.
Family A = ₹1940
Family B = ₹1750
Family C = ₹950
Answer: Family C.
2. Which family spent the most on healthcare?
Family A = ₹1200
Family B = ₹1280
Family C = ₹1400
Answer: Family C.
Family B = ₹1280
Family C = ₹1400
Answer: Family C.
3. Approximately how much did Family A spend on education?
From the table, Family A spent ₹1770 on education.
Answer: Approximately ₹1770.
Answer: Approximately ₹1770.
4. Which category did Family B spend most on?
Family B’s largest expenditure is housing:
₹1750.
Answer: Housing.
₹1750.
Answer: Housing.
5. Did Family A spend more on food or on healthcare?
Food = ₹1700
Healthcare = ₹1200
Therefore, Family A spent more on food.
Answer: Food.
Healthcare = ₹1200
Therefore, Family A spent more on food.
Answer: Food.
6. Which family spent the least amount overall?
Family A total:
1940 + 1700 + 1280 + 1200 + 1770 + 535 = ₹8425
Family B total:
1750 + 1546 + 1500 + 1280 + 1210 + 0 = ₹7286
Family C total:
950 + 1700 + 1540 + 1400 + 1300 + 150 = ₹7040
Answer: Family C.
1940 + 1700 + 1280 + 1200 + 1770 + 535 = ₹8425
Family B total:
1750 + 1546 + 1500 + 1280 + 1210 + 0 = ₹7286
Family C total:
950 + 1700 + 1540 + 1400 + 1300 + 150 = ₹7040
Answer: Family C.
7. Approximately how much did Family C spend overall?
Family C total:
950 + 1700 + 1540 + 1400 + 1300 + 150
= ₹7040
Answer: Approximately ₹7040.
950 + 1700 + 1540 + 1400 + 1300 + 150
= ₹7040
Answer: Approximately ₹7040.
8. Approximately what percentage of Family C’s expenses were on housing?
Housing = ₹950
Total = ₹7040
Percentage:
950/7040 × 100
≈ 13.5%
Answer: Approximately 13.5%.
Total = ₹7040
Percentage:
950/7040 × 100
≈ 13.5%
Answer: Approximately 13.5%.
Important:
Cluster-column charts are useful when we want to compare individual categories or groups. Stacked charts are useful when we also want to compare totals.
Stacked Bar Chart
In a stacked bar chart, the complete length of a bar represents the total. The smaller sections represent the components of that total.
IUCN Red List – Stacked Column Chart
Book Questions – Section 10.4
1. the following stacked coloumn chart shows the number of animal species in the IUCN red list , by class,over time.
(i) What does the number 14,234 on the top of the column 2019 represent?
The number 14,234 represents the total number of animal species shown in the IUCN Red List for the year 2019, across the classes/categories represented in the chart.
Answer: The total represented animal species in the 2019 column.
Answer: The total represented animal species in the 2019 column.
(ii) Approximately how many reptile species were in the list in 2016?
in 2016, the reptile section of the column is approximately 1500 reptile species.
Answer: approximately 1500 species
Answer: approximately 1500 species
(iii) Which class(es) of species have seen a relatively small increase in count between 2007 and 2019?
the classes that show a relatively small increase between 2007 and 2019 are; reptiles, amphibians and others.
Answer: reptiles, amphibians and others.
Answer: reptiles, amphibians and others.
2. The wickets taken by a bowler in international cricket matches till 2025are shwn in the table. complete the given stacked bar charts. (the bar lengths can be approximate.)
i) comparing the total wickets taken at home vs. overseas:
i) Home:62+100+32=194
So, total wickets at home:=194
Overseas:172+49+71=292
So, total wickets overseas: 292
For the stacked bar chart, draw:
Home: Test = 62, then ODI = 100, then T20 = 32 → total 194
Overseas: Test = 172, then ODI = 49, then T20 = 71 → total 292
Thus, the two bars should end approximately at 194 and 292 respectively.
So, total wickets at home:=194
Overseas:172+49+71=292
So, total wickets overseas: 292
For the stacked bar chart, draw:
Home: Test = 62, then ODI = 100, then T20 = 32 → total 194
Overseas: Test = 172, then ODI = 49, then T20 = 71 → total 292
Thus, the two bars should end approximately at 194 and 292 respectively.
ii) Total wickets taken in each format
Now add Home + Overseas for each format.
Test 62+172=234
ODI 100+49=149
T20 32+71=103
For the second stacked bar chart:
Test: Home = 62, Overseas = 172 → total 234
ODI: Home = 100, Overseas = 49 → total 149
T20: Home = 32, Overseas = 71 → total 103
10.2.2 An Alternative to Pie Chart
100% Stacked Bar Chart:
Every complete bar represents 100% of the total.
The individual sections show the percentage share of each category.
Percentage share =
(Category value ÷ Total value) × 100
Which family has the smallest share of their total expenditure towards healthcare?
Healthcare shares are:
Family A: 1200/8425 × 100 ≈ 14.24%
Family B: 1280/7286 × 100 ≈ 17.57%
Family C: 1400/7040 × 100 ≈ 19.89%
Therefore, the smallest share is for Family A.
Family A: 1200/8425 × 100 ≈ 14.24%
Family B: 1280/7286 × 100 ≈ 17.57%
Family C: 1400/7040 × 100 ≈ 19.89%
Therefore, the smallest share is for Family A.
Very Important:
A 100% stacked bar chart tells us about proportions, not actual totals.
If one bar has 60% of something and another has 40%, we cannot know which has more actual objects unless we know their total quantities.
Think and Reflect
1. What does this say about the scope of the stacked bars and 100% stacked bars?
A normal stacked bar chart shows both the total and the parts.
A 100% stacked bar chart mainly shows the percentage or proportion of each part.
2. Given a stacked bar chart can we make a corresponding 100% stacked bar chart?
Yes.
Calculate each component as a percentage of its total.
Then represent each total as a bar of length 100%.
3. Given a 100% stacked bar chart can we make a corresponding stacked bar chart?
Not from percentages alone.
We also need the actual total quantity for each bar.
4. What kind of inferences or comparisons can be made from a stacked bar chart and in a 100% stacked bar chart?
A stacked bar chart allows us to compare:
- total quantities,
- individual components, and
- parts of each total.
- percentage shares,
- proportions within each total, and
- how the composition of different totals differs.
A Typical Day
A 100% stacked bar chart can also represent how a fixed total time of 24 hours is divided among different activities.
Book Exercise – Section 10.5
1. The smart watch data of 5 people was tracked from Monday to Friday. The recorded data tells us how many hours each person spent lying down, sitting, and standing or moving around, etc., as shown in the following table.
(i) How much time does Sahana spend per day in each of these body-states? Make a reasonable guess and fill her row in the table.
(ii) Guess what activity or work Julie could be engaged in based on the data. Find out what your classmates’ guesses are.
(iii) Complete the chart below based on the tabular data.
(i)
The question asks for a reasonable guess. One reasonable choice is: Lying down = 8 hours
Sitting = 6 hours
Standing or moving around = 10 hours
Check: 8+6+10=24 hrs.
So Sahana’s row can be filled as:
Lying down 8 hrs.
Sitting 6hrs.
Standing or Moving around etc. 10 hrs.
(ii)
Julie spends:
8 hours lying down
10 hours sitting
6 hours standing or moving around
Since she spends a large amount of time sitting and some time moving around, she could be engaged in office or desk-based work, such as an office worker, accountant, or computer professional.
This is only a reasonable guess based on the data. julie could be an office/desk based worker.
(iii)
The question asks for a reasonable guess. One reasonable choice is: Lying down = 8 hours
Sitting = 6 hours
Standing or moving around = 10 hours
Check: 8+6+10=24 hrs.
So Sahana’s row can be filled as:
Lying down 8 hrs.
Sitting 6hrs.
Standing or Moving around etc. 10 hrs.
(ii)
Julie spends:
8 hours lying down
10 hours sitting
6 hours standing or moving around
Since she spends a large amount of time sitting and some time moving around, she could be engaged in office or desk-based work, such as an office worker, accountant, or computer professional.
This is only a reasonable guess based on the data. julie could be an office/desk based worker.
(iii)
Think 1. Will the bars look similar if a different teacher’s data is considered instead of Zakir’s?
Not necessarily.
Different teachers may spend different amounts of time sitting, standing and moving.
Therefore, their bars can have different proportions.
2. Will the bars look different if the data for all 7 days of the week is considered?
Yes, they may look different because Saturday and Sunday may have different activity patterns from weekdays.
3. Some bars in the chart may also match people doing other kinds of work/activities. Can you think of any? Discuss.
Yes.
For example, a bar with a large sitting portion could represent an office worker, student, driver or person working at a desk.
A bar with much more standing/moving could represent a shopkeeper, nurse, sportsperson or construction worker.
End-of-Chapter Exercises
Cricket Run Rate
1. In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6 runs in the first over making the run rate 6. (i) In the second over they scored 12 runs. What is the run rate now? (ii) After Over 19, their run rate was 6. What is the run rate after 20 overs given the team made 12 runs in the last over?
(i)
Total runs = 6 + 12 = 18
Overs = 2
Run rate = 18 ÷ 2 = 9 runs per over.
(ii)
After 19 overs, runs = 19 × 6 = 114.
Last over = 12 runs.
Total = 126 runs.
Run rate = 126 ÷ 20 = 6.3 runs per over.
Total runs = 6 + 12 = 18
Overs = 2
Run rate = 18 ÷ 2 = 9 runs per over.
(ii)
After 19 overs, runs = 19 × 6 = 114.
Last over = 12 runs.
Total = 126 runs.
Run rate = 126 ÷ 20 = 6.3 runs per over.
2. Five friends who play badminton surveyed the city and collected the price of shuttlecocks across brands and types (Nylon and Feather). The following is the list of the prices, in rupees (₹), that each one compiled.
Yusuf: 40(N), 105(N), 383(F), 108(N), 165(F), 116(F)
Srikanth: 194(N), 85(N), 93(N), 121(N)
Kashvi: 49(N), 297(F), 105(N), 275(F), 40(N)
Prasanna: 124(N), 333(F), 182(N), 258(F)
Gracy: 220(F), 458(F), 129(F), 183(N)
Can you describe a way they can work together to find the average price of a shuttlecock across types?
(i) Each one calculated the average of the prices they had gathered. Suppose these are a₁, a₂, a₃, a₄, a₅ respectively. Write an expression that gives the combined average.
(ii) Suppose aₙ and aₓ are the average prices of the nylon shuttlecocks and the feather shuttlecocks, respectively. Write an expression that gives the average price of a shuttlecock. Will this be equal to the answer we got using the method in the previous part?
Srikanth: 194(N), 85(N), 93(N), 121(N)
Kashvi: 49(N), 297(F), 105(N), 275(F), 40(N)
Prasanna: 124(N), 333(F), 182(N), 258(F)
Gracy: 220(F), 458(F), 129(F), 183(N)
Can you describe a way they can work together to find the average price of a shuttlecock across types?
(i) Each one calculated the average of the prices they had gathered. Suppose these are a₁, a₂, a₃, a₄, a₅ respectively. Write an expression that gives the combined average.
(ii) Suppose aₙ and aₓ are the average prices of the nylon shuttlecocks and the feather shuttlecocks, respectively. Write an expression that gives the average price of a shuttlecock. Will this be equal to the answer we got using the method in the previous part?
First method:
The five individual averages cannot simply be added and divided by 5 unless all five friends collected the same number of prices.
Yusuf collected 6 prices.
Srikanth collected 4 prices.
Kashvi collected 5 prices.
Prasanna collected 4 prices.
Gracy collected 4 prices.
Total number of prices = 6 + 4 + 5 + 4 + 4 = 23
the numbers of observations collected are 6, 4, 5, 4 and 4, then:
Combined average = (6a₁ + 4a₂ + 5a₃ + 4a₄ + 4a₅) ÷ 23
(ii)
Number of nylon shuttlecocks = 13
Number of feather shuttlecocks = 10
Total number of shuttlecocks = 23
Therefore, the average price is:
Average = (13aₙ+ 10aₓ) ÷ 23
Yes, this will be equal to the answer obtained in part (i), because both methods consider all 23 prices. Overall average price: 4063 / 23 = 176.65
Therefore, the average price of a shuttlecock is approximately: ₹176.65
The five individual averages cannot simply be added and divided by 5 unless all five friends collected the same number of prices.
Yusuf collected 6 prices.
Srikanth collected 4 prices.
Kashvi collected 5 prices.
Prasanna collected 4 prices.
Gracy collected 4 prices.
Total number of prices = 6 + 4 + 5 + 4 + 4 = 23
the numbers of observations collected are 6, 4, 5, 4 and 4, then:
Combined average = (6a₁ + 4a₂ + 5a₃ + 4a₄ + 4a₅) ÷ 23
(ii)
Number of nylon shuttlecocks = 13
Number of feather shuttlecocks = 10
Total number of shuttlecocks = 23
Therefore, the average price is:
Average = (13aₙ+ 10aₓ) ÷ 23
Yes, this will be equal to the answer obtained in part (i), because both methods consider all 23 prices. Overall average price: 4063 / 23 = 176.65
Therefore, the average price of a shuttlecock is approximately: ₹176.65
3. Shreyas holds 25 shares of a company at an average price of ₹150 and Vaishnavi holds 5 shares of the same company at an average price of ₹150. (i) Shreyas buys 10 shares of this company at a price of ₹30 each. What is the average price per share for him after the purchase? (ii) Vaishnavi buys some shares of this company at a price of ₹30 each and the average price per share for her after the purchase is ₹70. How many shares did she buy?
(i)
Original value = 25 × ₹150 = ₹3750
New purchase = 10 × ₹30 = ₹300
Total value = ₹4050
Total shares = 35
Average = 4050 ÷ 35
≈ ₹115.71
(ii)
Original value = 5 × 150 = ₹750
Let the number of new shares be x.
New value = 30x
New total shares = 5+x
Average is ₹70:
(750 + 30x)/(5+x) = 70
750 + 30x = 350 + 70x
400 = 40x
x = 10
Answer: 10 shares.
Original value = 25 × ₹150 = ₹3750
New purchase = 10 × ₹30 = ₹300
Total value = ₹4050
Total shares = 35
Average = 4050 ÷ 35
≈ ₹115.71
(ii)
Original value = 5 × 150 = ₹750
Let the number of new shares be x.
New value = 30x
New total shares = 5+x
Average is ₹70:
(750 + 30x)/(5+x) = 70
750 + 30x = 350 + 70x
400 = 40x
x = 10
Answer: 10 shares.
4. (Prithudakasvami commentary on Brahmagupta’s Brahmasphutasiddhanta, c. 864 CE) A “hasta” (meaning “forearm”) refers to a unit of length measuring around 18 inches. A pool 30 hastas long is dug to different depths along its length. It is divided into 5 sections having lengths 4, 5, 6, 7, and 8 hastas, and is dug to depths of 9, 7, 7, 3, and 2 hastas, respectively. Find the mean depth of the pool.
Use the lengths as weights because each depth applies to a different length.
Mean depth = (4×9 + 5×7 + 6×7 + 7×3 + 8×2) ÷ (4+5+6+7+8)
= (36 + 35 + 42 + 21 + 16) ÷ 30
= 150 ÷ 30
= 5 hastas
Answer: 5 hastas.
Mean depth = (4×9 + 5×7 + 6×7 + 7×3 + 8×2) ÷ (4+5+6+7+8)
= (36 + 35 + 42 + 21 + 16) ÷ 30
= 150 ÷ 30
= 5 hastas
Answer: 5 hastas.
5. Suvarna had purchased 1g gold at ₹15k (short for ₹15,000). This is shown by point O denoting the average price of gold that she possesses. Six different scenarios are given below for her next transaction. For each scenario estimate and mark the average price of gold she will have after the transaction.
(i) Purchase 1g gold at ₹30k
(ii) Purchase 2g gold at ₹30k
(iii) Purchase 1g gold at ₹10k
(iv) Purchase 10g gold at ₹10k
(v) Purchase 0.5g gold at ₹30k
(vi) Purchase 0.5g gold at ₹15k
Assume the original 1 g costs ₹15,000, so the original average is ₹15k per gram.
(i)
(15 + 30) ÷ 2 = ₹22.5k
(ii)
(15 + 2×30) ÷ 3
= 75/3
= ₹25k
(iii)
(15 + 10) ÷ 2
= ₹12.5k
(iv)
(15 + 10×10) ÷ 11
= 115/11
≈ ₹10.45k
(v)
(15 + 0.5×30) ÷ 1.5
= 30/1.5
= ₹20k
(vi)
Both purchases are at ₹15k per gram.
Therefore the average remains ₹15k.
(i)
(15 + 30) ÷ 2 = ₹22.5k
(ii)
(15 + 2×30) ÷ 3
= 75/3
= ₹25k
(iii)
(15 + 10) ÷ 2
= ₹12.5k
(iv)
(15 + 10×10) ÷ 11
= 115/11
≈ ₹10.45k
(v)
(15 + 0.5×30) ÷ 1.5
= 30/1.5
= ₹20k
(vi)
Both purchases are at ₹15k per gram.
Therefore the average remains ₹15k.
6. Given some data with corresponding weights, how would the weighted average change if all the weights are doubled? If required, experiment with some data. What do you observe? Justify your answer using algebra.
Suppose the weighted average is:
W = (w₁x₁ + w₂x₂ + … + wₙxₙ)/(w₁+w₂+…+wₙ)
If every weight is doubled:
W’ = (2w₁x₁ + 2w₂x₂ + … + 2wₙxₙ)/ (2w₁+2w₂+…+2wₙ)
Taking 2 common from numerator and denominator:
W’ = 2N/2D = N/D = W.
Answer: The weighted average does not change.
W = (w₁x₁ + w₂x₂ + … + wₙxₙ)/(w₁+w₂+…+wₙ)
If every weight is doubled:
W’ = (2w₁x₁ + 2w₂x₂ + … + 2wₙxₙ)/ (2w₁+2w₂+…+2wₙ)
Taking 2 common from numerator and denominator:
W’ = 2N/2D = N/D = W.
Answer: The weighted average does not change.
7. Answer the following questions based on the graph. (i) In 2003, approximately how many total objects were found orbiting Earth in space? In what year did this number double? (ii) Find the approximate number and share of payload objects and other objects in the year 2025. (iii) What can you say about the number of payload objects over time, and the number of other objects over time? What can you say about the share of payload objects over time, and the share of other objects over time?
The exact numerical answers depend on the graph accompanying the question, which is not included in the pasted text.
(i)From the graph, in 2003 the total number of objects was approximately:10,000
Double of 10,000 is:20,000
The graph reaches approximately 20,000 objects around 2021.
Answer: about 10,000 objects in 2003. it doubled around 2021.
(ii) rom the graph for 2025:
Payload objects ≈ 17,500
Other objects ≈ 15,500
Total ≈ 33,000
Share of payload objects; (17,000/33,000)×100 ≈ 53%
Share of other objects (15,500/33,000)×100 ≈ 47%
(iii) Number of payload objects: The number of payload objects increases over time. The increase becomes particularly rapid after about 2020.
Number of other objects:
The number of other objects also increases over time, but its increase is more gradual compared with payload objects.
Shares over time:
Earlier, other objects made up a larger share of the total. In recent years, the share of payload objects has increased considerably, while the share of other objects has decreased relatively.
Answer: Exact values require the textbook graph.
(i)From the graph, in 2003 the total number of objects was approximately:10,000
Double of 10,000 is:20,000
The graph reaches approximately 20,000 objects around 2021.
Answer: about 10,000 objects in 2003. it doubled around 2021.
(ii) rom the graph for 2025:
Payload objects ≈ 17,500
Other objects ≈ 15,500
Total ≈ 33,000
Share of payload objects; (17,000/33,000)×100 ≈ 53%
Share of other objects (15,500/33,000)×100 ≈ 47%
(iii) Number of payload objects: The number of payload objects increases over time. The increase becomes particularly rapid after about 2020.
Number of other objects:
The number of other objects also increases over time, but its increase is more gradual compared with payload objects.
Shares over time:
Earlier, other objects made up a larger share of the total. In recent years, the share of payload objects has increased considerably, while the share of other objects has decreased relatively.
Answer: Exact values require the textbook graph.
8. Observe the following infographic. What Percentage of Schools have a Playground?
(i) Identify the correct inference(s) from the statements given.
(a) More schools have a playground in Haryana compared to Uttarakhand.
(b) Approximately every 2 out of 3 schools in Arunachal Pradesh have a playground.
(c) Punjab has the highest number of schools with a playground.
(d) Suppose it is given that Maharashtra has more schools than Telangana. Then the number of schools having a playground is more in Maharashtra.
(ii) Using the information given, can we find the nation-wide percentage of schools with a playground? If not, what additional information is needed?
(i)
The map gives the percentage of schools having a playground in each state/UT. It does not give the actual number of schools.
(a) More schools have a playground in Haryana compared to Uttarakhand.
The map gives percentages, not the total number of schools. Therefore, we cannot compare the actual number of schools having playgrounds.
Incorrect.
(b) Approximately every 2 out of 3 schools in Arunachal Pradesh have a playground.
Arunachal Pradesh has approximately 48% of schools with playgrounds, not about 67%.
Incorrect.
(c) Punjab has the highest number of schools with a playground.
Again, only percentages are given. The total number of schools in each state is not given.
Incorrect.
(d) Suppose it is given that Maharashtra has more schools than Telangana. Then the number of schools having a playground is more in Maharashtra.
Maharashtra has about 93% of schools with playgrounds, while Telangana has about 74%. Since Maharashtra also has more schools, the number of schools with playgrounds must be greater in Maharashtra. Correct.
Answer: only d
(ii)
No, we cannot find it from the given information alone.
The map gives the percentage for each state, but we do not know the total number of schools in each state.
To calculate the nation-wide percentage, we would need the number of schools in each state/UT.
The calculation would be: Nation-wide percentage = (Total number of schools having playgrounds / Total number of schools) × 100
So, the additional information needed is:The total number of schools in each state/UT.
The map gives the percentage of schools having a playground in each state/UT. It does not give the actual number of schools.
(a) More schools have a playground in Haryana compared to Uttarakhand.
The map gives percentages, not the total number of schools. Therefore, we cannot compare the actual number of schools having playgrounds.
Incorrect.
(b) Approximately every 2 out of 3 schools in Arunachal Pradesh have a playground.
Arunachal Pradesh has approximately 48% of schools with playgrounds, not about 67%.
Incorrect.
(c) Punjab has the highest number of schools with a playground.
Again, only percentages are given. The total number of schools in each state is not given.
Incorrect.
(d) Suppose it is given that Maharashtra has more schools than Telangana. Then the number of schools having a playground is more in Maharashtra.
Maharashtra has about 93% of schools with playgrounds, while Telangana has about 74%. Since Maharashtra also has more schools, the number of schools with playgrounds must be greater in Maharashtra. Correct.
Answer: only d
(ii)
No, we cannot find it from the given information alone.
The map gives the percentage for each state, but we do not know the total number of schools in each state.
To calculate the nation-wide percentage, we would need the number of schools in each state/UT.
The calculation would be: Nation-wide percentage = (Total number of schools having playgrounds / Total number of schools) × 100
So, the additional information needed is:The total number of schools in each state/UT.
9. Decision Dilemma
(i) Which of the plays would you choose to watch based on the following chart?
i)The first chart is a 100% stacked bar chart. It shows the share/proportion of ratings for each play, but it does not show the total number of ratings.
From the chart, Play A has a large proportion of 4-star and 5-star ratings and a relatively small proportion of low ratings.
So, based only on the distribution of ratings, one could choose:Answer: Play A.
From the chart, Play A has a large proportion of 4-star and 5-star ratings and a relatively small proportion of low ratings.
So, based only on the distribution of ratings, one could choose:Answer: Play A.
(ii) The corresponding stacked bar chart is shown below. Would you change your decision after looking at this chart? Why/Why not? Discuss.
yes, the second chart gives actual numbers of ratings, rather than just their percentages.
Play B has received far more ratings than Play A and Play C. Therefore, the first chart can be misleading if we consider only percentages, because all three bars in the first chart have been made the same length.
However, if we are choosing based specifically on the average rating, the proportions of the ratings should be considered. Play A still has a slightly higher average rating than Play B based on the approximate values in the chart.
Therefore, the two charts provide different information:
First chart: compares the distribution/share of ratings.
Second chart: compares the actual number of ratings.
Play B: has many more total ratings.
Play A: has a slightly higher average rating based on the approximate distribution.
Conclusion: The second chart may affect the decision because it shows that Play B has a much larger number of ratings, but the choice depends on whether we value the rating distribution or the number of ratings received.
Play B has received far more ratings than Play A and Play C. Therefore, the first chart can be misleading if we consider only percentages, because all three bars in the first chart have been made the same length.
However, if we are choosing based specifically on the average rating, the proportions of the ratings should be considered. Play A still has a slightly higher average rating than Play B based on the approximate values in the chart.
Therefore, the two charts provide different information:
First chart: compares the distribution/share of ratings.
Second chart: compares the actual number of ratings.
Play B: has many more total ratings.
Play A: has a slightly higher average rating based on the approximate distribution.
Conclusion: The second chart may affect the decision because it shows that Play B has a much larger number of ratings, but the choice depends on whether we value the rating distribution or the number of ratings received.
10. A triathlon is an endurance race consisting of swimming 3.8 km, cycling 180 km, and running 42.2 km. Athletes compete for the fastest overall time, completing each segment in that order. The following table shows the finish time of 3 athletes in each segment.
(i) What is a suitable representation of this data-a stacked bar chart or a 100% stacked bar chart? Why do you think so?
(ii) Suppose a 100% stacked bar chart is drawn. Which of the following question(s) can be answered by looking at just that chart?
(a) Who finished the race first?
(b) Approximately what fraction of their race time did Athlete 1 spend cycling?
(c) Which athlete took the longest for running?
(i) What is a suitable representation of this data-a stacked bar chart or a 100% stacked bar chart? Why do you think so?
(ii) Suppose a 100% stacked bar chart is drawn. Which of the following question(s) can be answered by looking at just that chart?
(a) Who finished the race first?
(b) Approximately what fraction of their race time did Athlete 1 spend cycling?
(c) Which athlete took the longest for running?
(i)
A 100% stacked bar chart is more suitable.
Reason: The total race time is different for the three athletes. A 100% stacked bar chart makes each athlete’s total equal to 100%, so we can compare the fraction/percentage of time spent in swimming, cycling and running.
(ii)
A 100% stacked chart shows fractions of total race time. Therefore:
(a) Who finished first? A 100% stacked bar chart shows proportions, not the actual total time. Although Athlete 1’s total time is actually the least (9 h 23 min), this cannot be determined from the 100% chart alone.
No.
(b) Fraction of race time spent cycling? Yes.
(c) Who took the longest actual time for running? No.
Answer: Only (b).
A 100% stacked bar chart is more suitable.
Reason: The total race time is different for the three athletes. A 100% stacked bar chart makes each athlete’s total equal to 100%, so we can compare the fraction/percentage of time spent in swimming, cycling and running.
(ii)
A 100% stacked chart shows fractions of total race time. Therefore:
(a) Who finished first? A 100% stacked bar chart shows proportions, not the actual total time. Although Athlete 1’s total time is actually the least (9 h 23 min), this cannot be determined from the 100% chart alone.
No.
(b) Fraction of race time spent cycling? Yes.
(c) Who took the longest actual time for running? No.
Answer: Only (b).
11. Look at the following graph. What do you notice? What do you wonder? Write your inferences. Distribution of disabled persons by age group and type of disability in India (Census 2011)
What do I notice?
1) The graph divides people into four age groups: 0–19, 20–39, 40–59, and 60 years & above. Movement disability forms a significant portion in all age groups.
2) In the 60 years & above group, the proportions of seeing, hearing and movement disabilities appear relatively large.
3) Speech disability occupies a smaller portion compared with seeing, hearing and movement disabilities.
4) Multiple disabilities are present in every age group.
5) The proportion of seeing disability is noticeably higher in the 60+ group than in the younger groups. The proportion of movement disability is also quite large, particularly among the older age groups.
What do I wonder?
1) Why do seeing and movement disabilities become more prominent with increasing age?
2) Why is speech disability a relatively small proportion in all age groups?
3) What causes the differences in types of disabilities among different age groups?
4) How have these figures changed since the 2011 Census?
Inferences
1) Age appears to be associated with the distribution of different types of disabilities, particularly seeing and movement disabilities.
2) Older people have a relatively larger share of seeing and movement disabilities.
3) Disability is not of one type; there is considerable variation in the type of disability across age groups.
4) Multiple disabilities occur in all four age groups. The graph shows proportions within each age group, not the actual total number of disabled persons in each age group.
1) The graph divides people into four age groups: 0–19, 20–39, 40–59, and 60 years & above. Movement disability forms a significant portion in all age groups.
2) In the 60 years & above group, the proportions of seeing, hearing and movement disabilities appear relatively large.
3) Speech disability occupies a smaller portion compared with seeing, hearing and movement disabilities.
4) Multiple disabilities are present in every age group.
5) The proportion of seeing disability is noticeably higher in the 60+ group than in the younger groups. The proportion of movement disability is also quite large, particularly among the older age groups.
What do I wonder?
1) Why do seeing and movement disabilities become more prominent with increasing age?
2) Why is speech disability a relatively small proportion in all age groups?
3) What causes the differences in types of disabilities among different age groups?
4) How have these figures changed since the 2011 Census?
Inferences
1) Age appears to be associated with the distribution of different types of disabilities, particularly seeing and movement disabilities.
2) Older people have a relatively larger share of seeing and movement disabilities.
3) Disability is not of one type; there is considerable variation in the type of disability across age groups.
4) Multiple disabilities occur in all four age groups. The graph shows proportions within each age group, not the actual total number of disabled persons in each age group.
12. Maths Project. Do at least one of the following.
(i) Recall the previous day and fill in the approximate time spent lying down, sitting, and standing/moving around, etc. Ask at least 2 of your family members about their day and fill it in. Alternatively, you can choose to track just your own body-state for any one of the weekdays, Saturday, and Sunday. Visualise it using a stacked bar chart. Discuss with your class how you arrived at the estimated time spent in each state.
(i) Recall the previous day and fill in the approximate time spent lying down, sitting, and standing/moving around, etc. Ask at least 2 of your family members about their day and fill it in. Alternatively, you can choose to track just your own body-state for any one of the weekdays, Saturday, and Sunday. Visualise it using a stacked bar chart. Discuss with your class how you arrived at the estimated time spent in each state.
This is a data-collection activity, so there is no single numerical answer.
A good procedure is:
A good procedure is:
- Record the number of hours spent in each body-state.
- Make sure the total is approximately 24 hours.
- Draw one stacked bar for each person/day.
- Label each section.
- Compare the patterns and write observations.
(ii) Visualise your family’s monthly expenditure using a 100% stacked bar chart.
First divide expenditure into categories such as:
Percentage = category expenditure ÷ total expenditure × 100.
Then draw a 100% stacked bar for each month.
- Food
- Housing
- Education
- Transport
- Healthcare
- Entertainment
- Other expenses
Percentage = category expenditure ÷ total expenditure × 100.
Then draw a 100% stacked bar for each month.
(a) First, identify the expense categories in your family budget.
Possible categories are food, housing, education, transportation, healthcare, electricity, communication and recreation.
Choose categories that actually occur in the family budget.
(b) Discuss with your family members to decide how you will collect and organise the data.
Collect the amount spent in each category for each month.
Keep the data in a table so that totals and percentages can be calculated easily.
(c) Collect expense data for at least 3 months.
This is a practical activity.
Record the actual monthly expenditure for at least three months.
(d) Represent the data in a 100% stacked bar chart, with each bar showing how the total monthly expenditure is divided across the different categories.
Convert every category into a percentage of that month’s total and draw one 100% stacked bar for each month.
All bars must have the same total length because each represents 100%.
(e) Finally, write your observations and inferences based on the chart.
Possible observations:
- Which category takes the largest share.
- Which category takes the smallest share.
- Whether the shares change from month to month.
- Which expenses remain almost constant.
- Which expenses vary significantly.
Small-Group Project
13.Small-group project: Make a group of 3-4 students. Choose one scenario to design a custom rating scheme by assigning appropriate weights: (a) shopping experience at a cloth store, (b) travel experience in a bus, (c) tourism experience of a nearby tourist spot, (d) clinic/hospital experience.
Example: For a bus travel experience, possible aspects are:
- Comfort – weight 4
- Cleanliness – weight 3
- Safety – weight 5
- Timeliness – weight 3
(i) Discuss and arrive at 4-5 aspects to rate. For each aspect chosen, justify why it matters for an overall experience.
Choose aspects that genuinely affect the experience.
For example, in tourism: cleanliness, safety, facilities, scenery and accessibility.
Explain why each aspect matters.
(ii) Discuss and decide the relative weights and justify.
Give larger weights to aspects that are more important.
For example, safety may receive a higher weight than decoration.
(iii) Collect or imagine ratings from at least 10 people on a scale of 1-5 (5 being very good).
Prepare a table containing the ratings of the 10 people.
Each rating should be between 1 and 5.
(iv) Compute each individual rating. Compute the overall average across individuals.
For each person:
Weighted rating = Σ(weight × rating) ÷ Σ weights
Then find the average of the 10 individual weighted ratings.
Weighted rating = Σ(weight × rating) ÷ Σ weights
Then find the average of the 10 individual weighted ratings.
(v) Make a 100% stacked bar chart using your data.
Convert each weighted component into its percentage of the total weighted rating and show the components in a 100% stacked bar chart.
(vi) Write a short note on the observations and inferences, what your rating system captures well and what it misses.
Explain:
- Which aspect received the highest contribution.
- Which aspect received the lowest contribution.
- Whether the rating system gives a realistic overall picture.
- Which important factors may not have been included.
Whole Class Project
14.Whole class project: Each student shares the average age of their family and the number of family members. Discuss among the class and come up with a way to find out the average age of all the families of the class.
Do not simply average the family averages unless every family has the same number of members.
If family i has average age aᵢ and nᵢ members, its total age is aᵢnᵢ.
Therefore:
Overall average = Σ(aᵢnᵢ) ÷ Σnᵢ
This is a weighted average.
If family i has average age aᵢ and nᵢ members, its total age is aᵢnᵢ.
Therefore:
Overall average = Σ(aᵢnᵢ) ÷ Σnᵢ
This is a weighted average.
15.Given some data with corresponding weights, what would happen to the weighted average if all the weights are increased by a constant value, say 1? If required, experiment with some data. What do you observe? Justify your answer using algebra.
Unlike multiplying all weights by the same number, adding 1 to every weight generally changes the weighted average.
Original:
W = Σwᵢxᵢ / Σwᵢ
New:
W’ = Σ(wᵢ+1)xᵢ / Σ(wᵢ+1)
= [Σwᵢxᵢ + Σxᵢ] / [Σwᵢ+n]
This is generally not equal to W.
Answer: Adding the same constant to every weight generally changes the weighted average.
Original:
W = Σwᵢxᵢ / Σwᵢ
New:
W’ = Σ(wᵢ+1)xᵢ / Σ(wᵢ+1)
= [Σwᵢxᵢ + Σxᵢ] / [Σwᵢ+n]
This is generally not equal to W.
Answer: Adding the same constant to every weight generally changes the weighted average.
Farm Animals – Shares of Population
16. A farm has some cows, sheep, and chickens. Last year the cows made up 60%, the sheep 25%, and the chickens 15%. There was a decrease in the number of all three animals’ population over the year. Choose the possibilities for the change in their respective shares of the population:
(i) % of cows decreased, % of sheep decreased, % of chickens decreased
(ii) % of cows increased, % of sheep increased, % of chickens increased
(iii) % of cows remained the same, % of sheep remained the same, % of chickens remained the same
(iv) % of cows decreased, % of sheep increased, % of chickens remained the same
(v) % of cows increased, % of sheep increased, % of chickens decreased.
(i) % of cows decreased, % of sheep decreased, % of chickens decreased
(ii) % of cows increased, % of sheep increased, % of chickens increased
(iii) % of cows remained the same, % of sheep remained the same, % of chickens remained the same
(iv) % of cows decreased, % of sheep increased, % of chickens remained the same
(v) % of cows increased, % of sheep increased, % of chickens decreased.
The shares always add up to 100%.
Therefore, they cannot all decrease or all increase at the same time.
(i) Not possible because all three shares cannot decrease and still total 100%.
(ii) Not possible because all three cannot increase and total 100%.
(iii) Possible. All three populations may decrease by the same proportion, so their shares remain unchanged.
(iv) Possible. One share can decrease, another can increase, while the third remains unchanged.
(v) Possible. Two shares can increase while one decreases, with the total remaining 100%.
Answer: (iii), (iv) and (v) are possible.
(i) Not possible because all three shares cannot decrease and still total 100%.
(ii) Not possible because all three cannot increase and total 100%.
(iii) Possible. All three populations may decrease by the same proportion, so their shares remain unchanged.
(iv) Possible. One share can decrease, another can increase, while the third remains unchanged.
(v) Possible. Two shares can increase while one decreases, with the total remaining 100%.
Answer: (iii), (iv) and (v) are possible.
Chapter Summary
Weighted Average / Weighted Mean
x̄ = (w₁x₁ + w₂x₂ + w₃x₃ + … + wₙxₙ) ÷ (w₁ + w₂ + w₃ + … + wₙ)
x̄ = (w₁x₁ + w₂x₂ + w₃x₃ + … + wₙxₙ) ÷ (w₁ + w₂ + w₃ + … + wₙ)
Key Points to Remember
- An average of averages must take the sizes of the groups into account.
- If all groups have equal sizes, their averages can be averaged directly.
- For mixtures, larger quantities have greater influence on the final concentration.
- A weighted average is used when different values have different importance.
- Weighted averages are useful for marks, ratings, mixtures, prices and many other situations.
- If all weights are multiplied by the same positive number, the weighted average does not change.
- If the same number is added to every weight, the weighted average generally changes.
- A stacked bar chart shows totals as well as their components.
- A 100% stacked bar chart shows proportions or percentages.
- A 100% stacked bar chart does not show actual totals unless additional information is provided.
- A stacked bar chart can be converted into a 100% stacked bar chart by calculating percentages.
- A 100% stacked bar chart cannot be converted into an ordinary stacked bar chart unless the actual totals are known.
Quick Revision
| Concept | Remember |
|---|---|
| Average | Total ÷ Number of values |
| Combined average | Total of all groups ÷ Total number of values |
| Weighted average | Σ(wx) ÷ Σw |
| Mixture concentration | Total substance ÷ Total mixture |
| Stacked bar chart | Shows totals and components |
| 100% stacked bar chart | Shows proportions/percentages |
Exam Tip:
Whenever you see words such as different group sizes, importance, weightage, concentration, quantity, contribution, rating, marks or purity, think about a weighted average.
prepare more with the given Worksheet, which will help you to improve your knowledge .
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